138 
Mathenati- 
eal Geogta- 
To find the 
angles of 
position be- 
tween two 
places. 
Solution of 
questions in 
navigation. 
Given diffe- 
rence of 
longitude 
and lati- 
tude, to 
find the 
course and 
distance 
sailed. 
Given dif- 
ference of 
latitude and 
course, to 
find diffe- 
rence of 
longitude 
and dis- 
tance, 
Given dif- 
ference of 
jongitude 
and course, 
to find the 
difference 
of latitude 
and dis- 
tance. 
Pros, XXII.—Given two places, it is required to 
find the angle which agreat circle passing through them, 
makes with the meridian of each. wiles 
When the places are under the same meridian, the 
angles are 0, and when they are under the equator the 
angles are 90°. é 6.7 
When the places are neither on the same meridian, 
nor under the equator, place the globe in such a posi- 
tion, that both the places may be in the horizon toge- 
ther, and a line drawn with a pencil along the horizon 
will be a great circle passing through the places ; the 
globe being then rectified for both places successively, 
as in Prob. I. and the places themselves brought to the 
brazen meridian, the arches of the horizon, intercepted 
between the meridian and the circle passing through 
the places, will be the measures of the angles required. 
This and the preceding problem may be very con- 
veniently applied to the solution of questions in globu- 
lar sailing. Thus, if the ship be steered on an arch of 
a great circle, the distance sailed is found by Prob. XXI. 
and the rhumbs on. which the ship must leave the:one 
place and arrive at the other, by Prob. XXIL. 
Example 1. A ship leaves A, Lat. 16° south, Long. 
5° 50’ west, and sailing on .a great circle arrives at B, 
Lat. 82° 30’ north, Long. 63° 30’ west, required the 
distance sailed, and the rhumbs on which she left A 
and arrived at B? 
1. Describe on the globe with a pencil a great circle 
passing through A pal B, and find the length of the 
intersected arch by Prob. XX. which in this case will 
be about 73° 30’, or 4410 geographical miles. 
2. Rectify the globe for the latitude of A, bring it to 
the meridian, and mark the point where the great circle 
passing through B intersects the horizon, which will 
be about 48° from the north point towards the west. 
8. Rectify the globe for the latitude of B; bring it 
to the meridian, and mark the. point of the horizon 
again intersected by the great circle, and which will 
now be found to be nearly 58° from the north towards 
the west: It appears, therefore, that a ship sailing from 
A to B, upon a great circle, must leave A on the rhumb 
N. 48° W., and gradually changing her course, must 
arrive at B on the rhumb N. 58° W. having run a dis- 
tance of about 4410 geographical. miles. 
— 2. A shipleaves A, Lat. 16° S..on a course 
N. 48° W, and sailing on a great circle arrives at B, 
Lat. 32° 30’ N.; required the difference of longitude and 
distance between A and B? 
Rectify the globe for the latitude of A, and bring it 
to the meridian. Fix the quadrant of altitude on A, 
and make its graduated edge intersect the horizon in 
the given course (N. 48° W.) ; then the point where the 
quadrant of altitude intersects the parallel of 32° 30’, is 
the position of B. The place being thus determined, 
the difference of longitude may be found by Prob. 
XVIII. and the distance by Prob. XXI. In this ex- 
ample the former is about 57° 40’, and the latter 4410 
geographical miles. 
Example 3. A ship sails from B, Lat. 32° 30’ N. on 
a course §, 58° E. and when she reaches A, her diffe- 
rence of longitude is 57° 40’ ; required the difference of 
latitude, and distance between A and B, the ship ha- 
ving sailed on a great circle? . 
Bring B to the meridian, and rectify the globe for its 
latitude. Place the quadrant of altitude on B, and let 
its graduated edge intersect the horizon in the given 
course (S. 58° tp oo the point where the quadrant 
intersects a meridian, 57°40’ E. of the meridian of B, 
will give the position of A, and A being determined, the 
5 
GEOGRAPHY.) 
-or setting, as‘well as ‘his amplitude, on 
difference of latitude and distance may be found by 
Probs. XVIII. erat The rap an de 
is 48° $0’, and the latter 4410 geographical miles, 
Examples of this kind rok. be multiplied, but the 
ing are sufficient to illustrate the general prin- 
iple. See Navication. iaving Dita ob 
Pros. XXIII. To find those places in the torrid To 
zone, to which the sun will be vertical, on any given ¥ 
da: tical 
Prob. II: sa day. 
mark 
Find the sun’s place in the ecliptic by 
lution 2d, bring that place to the meridian, and 
the degree of the meridian over it ; all the places that 
under that degree while the globe revolves, will 
Kiovéraiee sun vertical that day. ¢ CaNeTs iD bs 
Prox. XXIV.—The day and hour being given at To fir 
any ‘place, to find where'the sun is then vertical... where 
Find the sun’s declination, or the parallel to which he call di 
is vertical that day, and bring the bape lace.and hour jour, 
tothe meridian ; then turn the globe, till the 12th hour : 
at noon come tothe meridian, and the intersectionof —__ 
the meridian, with the parallel of latitude to ;which’the 
sun is vertical, will be the place required. au WF My ‘ 
Pros. XXV.—A place being given in the torrid To f 
zone, to find:on what days the sun will be vertical there, when 
Find the latitude of the and the points of the £"" 
ecliptic which have the same latitude; the days im the iv. 
ane og opposite to'these points, will be the days re- © 
‘ Pros. XXVI.—To find the sun’s altitude and. azimuth To fi 
at any given time and place. ge “ODS 
Rectify the globe for the given latitude, and bring tudes 
the sun’s place in the ecliptic and the 12th hour of the = 
horary to the brazen meridian. Turn the globe to- 
wards the east or west, according as the time is before 
or after mid-day, till the difference between the gi 
hour and 12 is under the meridian. Fix the pie wi 
of altitude on the zenith, and make its 
fall on the sun’s place in the ecliptic; 
of the quadrant over the sun’s place will be his altitude, 
and the point of the horizon intersected by the quadrant 
his azimuth. et 
If the sun’s meridian altitude ris sme it is found 
by rectifying the globe for the latitude, and bri 
the sun’s place to the meridian, when the arch of the 
meridian a between the sun’s place and the 
horizon will be the altitude required. In this case the 
azimuth is nothing. Rhy 
Pros. XXVII—To find the circle of illumination To 
for any day, and the places that are above or below it cite 
o any hour of that day, reckoning the time at a given ™ 
ace, 
i Rectify the globe for the latitude to which the sun is 
vertical, and the horizon will represent the circle of il<_ 
lumination. Bring the given and hour to the 
meridian, and turn the globe towards the west or east, 
according as the hour is before or after mid-day, till 
12 of the hour circle be under the meridian ; then the 
horizon will represent the circle of illumination at the 
ot hour. “To the places in the western side of the 
orizon, the sun is rising, to those in the easternside, heis 
setting, and to those under the meridian, he is culmi- 
nating. 
Pros. XXVIII.—To find the hour of the sun’s rising To 
any day, at any ho 
place, w latitude does not exceed 66° 32’. 
Rectify the globe for the latitude of the place, and “ 
bring the sun’s place in the ecliptic, and the 12th hour — 
of the horary, to the brazen meridian. Then turning — 
the globe eastward till the sun’s place be in the horizon, 
